Don't understand equation with overleftrightarrow symbol

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  • #1
joneall
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Equation in Srednicki QFT book
I've started reading Srednicki's book on QFT, which was starting well. Then I hit on an equation which I just don't understand at all. Since I don't know what the symbol is called, I can only refer to it by its latex name.
Here's the bit. Srednicki defines the following object:
$$f \overleftrightarrow{\partial_{\mu}}g := f (\partial_{\mu}{g}) - (\partial_{\mu}f ) g $$.
Already, it is not clear to me if the second term is a function of a derivative or a product.
He goes on by deriving
$$i \overleftrightarrow{\partial_{0}} \phi(x) =i \partial_0 \phi(x) + \omega \phi(x) $$
clearly using
$$ \partial_0 \phi = i \omega \phi $$.

I will probably feel like an idiot when someone explains this to me, but I just can't get it. How are those two equations compatible? Does this double-arrow beast have a name?
 
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  • #2
Page number or other pointer?
 
  • #3
The second term is a product. You first apply the differential operator to the left member of the original product of functions. You obtain a new function which you then multiply with the function which was/is at the right of the original product. Pay attention, the functions of spacetime may not commute in a product, so keep the exact order.
 
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  • #4
Nugatory said:
Page number or other pointer?
Srednicki, QFT. page 26, just after equation 3.21. Sorry I didn't mention that.
 
  • #5
It looks a bit like differentiation by parts, but with a minus sign.
 
  • #6
dextercioby said:
The second term is a product. You first apply the differential operator to the left member of the original product of functions. You obtain a new function which you then multiply with the function which was/is at the right of the original product. Pay attention, the functions of spacetime may not commute in a product, so keep the exact order.
Uh, sorry. I didn't get that. I'm wondering what the "beast" with the double arrow over it is. And what f is.
 
  • #7
I think here is the answer.
 
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  • #8
Aha, yes indeed. So it is sortuv like differentiation by parts of a product of functions, only with a minus sign. I'm guessing this has to do with the metric signature, which would explain why it could be the inverse.
Why oh why did people start using opposite metric signatures?
In any case, thanks to all.
 
  • #9
What Sredinicki evaluates in (3.21) using this symbol is
$$\begin{split}
\exp(-\mathrm{i} k x) \overleftrightarrow{\partial_0} \varphi(x)&=\exp(-\mathrm{i} k x) \partial_0 \varphi(x) - [\partial_0 \exp(-\mathrm{i} k x)] \varphi(x) \\
&= \exp(-\mathrm{i} k x) [\partial_0 \varphi(x) +\mathrm{i} k_0 \varphi(x)]\\
&=\exp(-\mathrm{i} k x) [\partial_0 \varphi(x) - \mathrm{i} \omega \varphi(x)].
\end{split}$$
In the last step I used that the four-momentum is on-shell, ##\omega=\sqrt{\vec{k}^2+m^2}##, and that Srednicky uses the (-+++) convention of the Lorentz fundamental form.
 
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  • #10
joneall said:
So it is sortuv like differentiation by parts of a product of functions, only with a minus sign. I'm guessing this has to do with the metric signature, which would explain why it could be the inverse.
Why oh why did people start using opposite metric signatures?
The minus sign has nothing to do with any particular convention for the signature of the 4D metric. Even in ordinary quantum mechanics, conservation of probability requires that the wavefunction ##\psi## of a particle of mass ##m## should satisfy:$$\frac{\partial}{\partial t}\left(\psi^{*}\psi\right)+\mathbf{\nabla}\cdot\mathbf{j}=0$$where the probability current density ##\mathbf{j}## is given by:$$\mathbf{j}\equiv-\frac{i\hslash}{2m}\left(\psi^{*}\mathbf{\nabla}\psi-\psi\mathbf{\nabla}\psi^{*}\right)=-\frac{i\hslash}{2m}\psi^{*}\mathbf{\overleftrightarrow{\nabla}}\psi$$So the double-arrow differential operator is useful even in a non-relativistic setting.
 
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  • #11
That ##k_0=-k^0=-\omega## depends on the sign convention of the Minkowski fundamental form. The physics is of course completely independent of the choice of convention.
 

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